we have
\(f(x)=6x^{2}+12x-7\)
Equate the function to zero
\(6x^{2}+12x-7=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(6x^{2}+12x=7\)
Factor the leading coefficient
\(6(x^{2}+2x)=7\)
Complete the square. Remember to balance the equation by adding the same constants to each side.
\(6(x^{2}+2x+1)=7+6\)
\(6(x^{2}+2x+1)=13\)
Rewrite as perfect squares
\(6(x+1)^{2}=13\)
\((x+1)^{2}=13/6\)
Square root both sides
\(x+1=(+/-)\sqrt{\frac{13}{6}}\)
\(x=-1(+/-)\sqrt{\frac{13}{6}}\)
\(x1=-1+\sqrt{\frac{13}{6}}\)
\(x2=-1-\sqrt{\frac{13}{6}}\)
therefore
the answer is
The zeros of the quadratic equation are
\(x1=-1+\sqrt{\frac{13}{6}}\)
\(x2=-1-\sqrt{\frac{13}{6}}\)
